(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Question:

$(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$ is equal to

(a) 0
(b) 1
(c) −1
(d) None of these

Solution:

The given expression is

$(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$

Simplifying the given expression, we have

$(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$

$=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$

$=\frac{1-\sin ^{2} \theta}{\sin \theta} \times \frac{1-\cos ^{2} \theta}{\cos \theta} \times \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}$

$=\frac{\cos ^{2} \theta}{\sin \theta} \times \frac{\sin ^{2} \theta}{\cos \theta} \times \frac{1}{\sin \theta \cos \theta}$

$=\frac{\cos ^{2} \theta \sin ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}$

$=1$

Therefore, the correct option is (b).